Question

If the force on a rocket, that releases the exhaust
gases with a velocity of 300 m/s is 210 N, then
the rate of combustion of the fuel is
(1) 0.07 kg/s
(2) 1.4 kg/s
(3) 0.7 kg/s
(4) 10.7 kg/s

Answer 1

Answer:

rate of change of momentum is force apply it ...and will be 0.7

Answer 2

Answer:

Force=d/dt (momentum)

=d/dt(mv)

=v(dm/dt)

=210=300(dm/dt)

=dm/dt(rate of combustion)

=210/300=0.7kg/s ans.

Explanat